Classical-Mechanics · Unit 20 · Video 3 · Interactive Practice

Rotational Kinetic Energy and Moment of Inertia: Deriving I for a Uniform Rod

IKey Formulas

Formula	Name	What it says
$K_{\mathrm{cm}} = \tfrac{1}{2}\,I_{\mathrm{cm}}\,\omega_{\mathrm{cm}}^{2}$	Rotational kinetic energy	$I_{\mathrm{cm}}$ stands where $m$ stands, $\omega_{\mathrm{cm}}$ where $v$ stands
$I_{\mathrm{cm}} = \displaystyle\int_{\mathrm{body}} r_{dm}^{2}\, dm$	Moment of inertia	Every mass element weighted by its distance squared; units $\mathrm{kg}\cdot\mathrm{m}^{2}$
$dm = \lambda\, dx, \quad \lambda = \dfrac{m}{L}$	One slice of a uniform rod	Constant mass per unit length turns $dm$ into $dx$ ; a slice at $x$ has $r_{dm}^{2} = x^{2}$
$I_{\mathrm{cm}} = \tfrac{1}{12}\, m L^{2}$	Uniform rod, axis through the centre	What the integral evaluates to

Key Insight: A rod spinning about its centre has $\tfrac{1}{2}m v_{\mathrm{cm}}^{2} = 0$ and yet carries kinetic energy. The mass of each element enters once and its distance from the axis enters squared, so mass far out counts for much more than mass near the axis, and mass on the axis contributes nothing at all — which is why $I$ belongs to a body and an axis, never to the body alone.

IIOne Rod, Many Speeds

The centre of mass never moves, yet every other slice does — at a speed set by its distance from the axis.

Rotation angle 25° ω (rad/s) 2.0 Slice at x (m) 0.6

IIIFrom a Sum of Slices to an Integral

Each slice contributes $\Delta m\, x^{2}$ ; refine the slicing and the sum settles onto $\int x^{2}\, dm$ .

Slices N 4

The same limit defines $I$ for any body about any axis; move this axis to one end of the same rod and the integral runs from $0$ to $L$ instead, returning a different number.

IVEvaluating the Integral for the Rod

One integral along $x$ turns $\lambda$ , $dx$ and $x^{2}$ into $\tfrac{1}{12}mL^{2}$ .

Step 1

Step 1 — one slice

\lambda = \frac{m}{L}, \qquad dm = \lambda\, dx

r_{dm} = |x| \;\Longrightarrow\; r_{dm}^{2} = x^{2}

Step 2 — integrate over the rod

I_{\mathrm{cm}} = \int_{\mathrm{body}} r_{dm}^{2}\, dm = \int_{-L/2}^{L/2} x^{2}\, \lambda\, dx

= \lambda \int_{-L/2}^{L/2} x^{2}\, dx

Step 3 — antiderivative

= \lambda\, \frac{x^{3}}{3}\, \Big|_{-L/2}^{\,L/2}

Step 4 — the two limits

= \frac{m}{L}\frac{(L/2)^{3}}{3} - \frac{m}{L}\frac{(-L/2)^{3}}{3}

= \frac{mL^{2}}{24} + \frac{mL^{2}}{24} = \tfrac{1}{12}\, m L^{2}

Step 5 — back to the energy

K_{\mathrm{cm}} = \tfrac{1}{2}\, I_{\mathrm{cm}}\, \omega_{\mathrm{cm}}^{2} = \tfrac{1}{2}\cdot\tfrac{1}{12}\, m L^{2} \omega_{\mathrm{cm}}^{2} = \tfrac{1}{24}\, m L^{2} \omega_{\mathrm{cm}}^{2}

VQuiz Questions

Problem 1 · Energy of a Spinning Rod

Given: a uniform rod of mass $m = 4.0\ \mathrm{kg}$ and length $L = 3.0\ \mathrm{m}$ spins at $\omega_{\mathrm{cm}} = 2.0\ \mathrm{rad/s}$ about a perpendicular axis through its centre — find its kinetic energy $K_{\mathrm{cm}}$ .

✅ Correct!

I_{\mathrm{cm}} = \tfrac{1}{12}(4.0)(3.0)^{2} = 3.0\ \mathrm{kg}\cdot\mathrm{m}^{2}

, so

K_{\mathrm{cm}} = \tfrac{1}{2}(3.0)(2.0)^{2} = 6.0\ \mathrm{J}

❌ The angular velocity is squared.

\tfrac{1}{2}(3.0)(2.0) = 3.0

uses

\omega

, not

\omega^{2}

;

\tfrac{1}{2}(3.0)(2.0)^{2} = 6.0\ \mathrm{J}

❌ That gives every gram the end speed.

\tfrac{1}{2}m(\omega L/2)^{2} = \tfrac{1}{2}(4.0)(3.0)^{2} = 18\ \mathrm{J}

assumes all the mass sits at

x = \pm L/2

; the integral averages

x^{2}

L^{2}/12

, not

L^{2}/4

❌ That is the wrong axis.

\tfrac{1}{3}mL^{2} = 12\ \mathrm{kg}\cdot\mathrm{m}^{2}

belongs to an axis through one end; through the centre the integral gives

\tfrac{1}{12}mL^{2} = 3.0\ \mathrm{kg}\cdot\mathrm{m}^{2}

Show solution

Step 1 — the moment of inertia. The axis is perpendicular to the rod through its centre of mass, so

I_{\mathrm{cm}} = \tfrac{1}{12}mL^{2} = \tfrac{1}{12}(4.0\ \mathrm{kg})(3.0\ \mathrm{m})^{2} = 3.0\ \mathrm{kg}\cdot\mathrm{m}^{2}

Step 2 — the energy. Every element shares the one angular velocity, so

K_{\mathrm{cm}} = \tfrac{1}{2}I_{\mathrm{cm}}\omega_{\mathrm{cm}}^{2} = \tfrac{1}{2}(3.0\ \mathrm{kg}\cdot\mathrm{m}^{2})(2.0\ \mathrm{rad/s})^{2} = 6.0\ \mathrm{J}

The centre of mass stays put, so the translational formula would have given $\tfrac{1}{2}mv_{\mathrm{cm}}^{2} = 0$ : all $6.0\ \mathrm{J}$ live in the rotation.

Problem 2 · The Lower Limit

Given: the rod integral $I_{\mathrm{cm}} = \lambda\displaystyle\int_{-L/2}^{L/2} x^{2}\, dx = \lambda\,\frac{x^{3}}{3}\Big|_{-L/2}^{\,L/2}$ , and a student who argues that because $(-L/2)^{3} = -(L/2)^{3}$ the two halves cancel — find what the lower limit actually contributes to $I_{\mathrm{cm}}$ .

✅ Correct! The antiderivative at the lower limit is

-\dfrac{mL^{2}}{24}

, and the evaluation subtracts it:

\dfrac{mL^{2}}{24} - \left(-\dfrac{mL^{2}}{24}\right) = \dfrac{mL^{2}}{12}

❌ That is the antiderivative, not the contribution.

\lambda\dfrac{(-L/2)^{3}}{3} = -\dfrac{mL^{2}}{24}

is what you evaluate at the lower limit; the fundamental theorem subtracts it, so it enters

I_{\mathrm{cm}}

with a plus sign.

❌ Nothing cancels here. Both halves are made of real mass at real distances, and

r_{dm}^{2} = x^{2} \ge 0

everywhere: a moment of inertia can never come out zero for a body with mass off the axis.

❌ That is the total, both limits together. The two halves contribute

\dfrac{mL^{2}}{24}

each, and only their sum is

\dfrac{mL^{2}}{12}

Show solution

Evaluate both limits. With $\lambda = m/L$ ,

\lambda\,\frac{(L/2)^{3}}{3} = \frac{m}{L}\cdot\frac{L^{3}}{24} = \frac{mL^{2}}{24}, \qquad \lambda\,\frac{(-L/2)^{3}}{3} = -\frac{mL^{2}}{24}

Subtract, do not add.

I_{\mathrm{cm}} = \frac{mL^{2}}{24} - \left(-\frac{mL^{2}}{24}\right) = \frac{mL^{2}}{24} + \frac{mL^{2}}{24} = \tfrac{1}{12}mL^{2}

The minus sign of the evaluation and the minus sign of the cube meet and produce a second positive term — exactly what symmetry demands: the two halves of a uniform rod are mirror images about the axis, so they must contribute equally.

The cancellation the student expected happens for the first moment $\displaystyle\int x\, dm$ , which really is zero about the centre of mass. Squaring the distance destroys that cancellation.

Problem 3 · Working Backwards from the Energy

Given: a uniform rod of mass $m = 3.0\ \mathrm{kg}$ and length $L = 1.2\ \mathrm{m}$ spins about a perpendicular axis through its centre with kinetic energy $K_{\mathrm{cm}} = 4.5\ \mathrm{J}$ — find the angular velocity and the speed of one end.

What is the angular velocity?

How fast does one end move?

✅ Correct!

I_{\mathrm{cm}} = 0.36\ \mathrm{kg}\cdot\mathrm{m}^{2}

gives

\omega_{\mathrm{cm}} = \sqrt{2K/I} = 5.0\ \mathrm{rad/s}

, and the end sits at

r = L/2 = 0.60\ \mathrm{m}

, so

v = \omega r = 3.0\ \mathrm{m/s}

❌ Mass cannot stand in for $I$ .

\sqrt{2K/m} = \sqrt{3.0} = 1.7

has units of

\mathrm{m/s}

, not

\mathrm{rad/s}

: the rotational formula needs

I_{\mathrm{cm}} = \tfrac{1}{12}mL^{2}

❌ That uses the end-axis value.

\tfrac{1}{3}mL^{2} = 1.44\ \mathrm{kg}\cdot\mathrm{m}^{2}

gives

2.5\ \mathrm{rad/s}

; the axis here runs through the centre, where

I_{\mathrm{cm}} = 0.36\ \mathrm{kg}\cdot\mathrm{m}^{2}

❌ That is $\omega^{2}$ .

2K/I = 25\ \mathrm{rad^{2}/s^{2}}

still has to be square-rooted.

❌ Check the inversion.

K = \tfrac{1}{2}I\omega^{2}

rearranges to

\omega = \sqrt{2K/I}

with

I = \tfrac{1}{12}mL^{2}

❌ The radius is half the rod. An end sits

L/2 = 0.60\ \mathrm{m}

from the axis, not

L = 1.2\ \mathrm{m}

, so

v = \omega L/2 = 3.0\ \mathrm{m/s}

❌ That is the radius, not the speed.

0.60\ \mathrm{m}

L/2

; multiply by

\omega_{\mathrm{cm}}

to get a speed.

❌ Only the centre is at rest.

v_{\mathrm{cm}} = 0

holds at

x = 0

; the speed of a slice is

|x|\omega

, largest at the ends.

❌ Check the radius. A slice at position

x

moves at

|x|\omega_{\mathrm{cm}}

, and an end has

|x| = L/2

Show solution

Step 1 — the moment of inertia.

I_{\mathrm{cm}} = \tfrac{1}{12}mL^{2} = \tfrac{1}{12}(3.0)(1.2)^{2} = 0.36\ \mathrm{kg}\cdot\mathrm{m}^{2}

Step 2 — invert the energy formula.

\omega_{\mathrm{cm}} = \sqrt{\frac{2K_{\mathrm{cm}}}{I_{\mathrm{cm}}}} = \sqrt{\frac{2(4.5)}{0.36}} = \sqrt{25} = 5.0\ \mathrm{rad/s}

Step 3 — the speed of an end. An end is a mass element at $|x| = L/2 = 0.60\ \mathrm{m}$ , and every element travels on a circle at $v = |x|\omega_{\mathrm{cm}}$ :

v = (0.60\ \mathrm{m})(5.0\ \mathrm{rad/s}) = 3.0\ \mathrm{m/s}

Check. If the whole rod moved at $3.0\ \mathrm{m/s}$ its energy would be $\tfrac{1}{2}(3.0)(3.0)^{2} = 13.5\ \mathrm{J}$ — three times the true $4.5\ \mathrm{J}$ , because the average of $x^{2}$ over the rod is $L^{2}/12$ , one third of the end value $L^{2}/4$ .

Problem 4 · Same Mass, Same Length, Different Body

Given: a uniform rod of mass $m$ and length $L$ , and a dumbbell made of two point masses $m/2$ joined by a massless bar of length $L$ , both rotating about a perpendicular axis through the centre — find the ratio $I_{\text{dumbbell}} / I_{\text{rod}}$ .

✅ Correct!

I_{\text{dumbbell}} = 2\left(\tfrac{m}{2}\right)\left(\tfrac{L}{2}\right)^{2} = \tfrac{1}{4}mL^{2}

, and

\tfrac{1}{4}mL^{2} \big/ \tfrac{1}{12}mL^{2} = 3

❌ Equal mass does not mean equal $I$ . Mass is one number for a body, but

I = \int r^{2} dm

also asks where that mass sits: all of the dumbbell's mass is at

L/2

, while most of the rod's is nearer the axis.

❌ The ratio is upside down. Moving mass outwards can only raise

I

, so the dumbbell — with every gram at the maximum distance — must have the larger moment of inertia.

❌ Check the masses. Each end carries

m/2

, not

m

2\left(\tfrac{m}{2}\right)\left(\tfrac{L}{2}\right)^{2} = \tfrac{1}{4}mL^{2}

, not

\tfrac{1}{2}mL^{2}

Show solution

The dumbbell. The integral collapses to a two-term sum, since all the mass sits at $|x| = L/2$ :

I_{\text{dumbbell}} = \sum_i \Delta m_i r_i^{2} = \frac{m}{2}\left(\frac{L}{2}\right)^{2} + \frac{m}{2}\left(\frac{L}{2}\right)^{2} = \frac{mL^{2}}{4}

The rod. Its mass is spread evenly along $x$ , and the integral gives

I_{\text{rod}} = \tfrac{1}{12}mL^{2}

The ratio.

\frac{I_{\text{dumbbell}}}{I_{\text{rod}}} = \frac{mL^{2}/4}{mL^{2}/12} = 3

Spun at the same $\omega$ , the dumbbell therefore stores three times the kinetic energy of the rod, with the same mass turning at the same rate. The squared distance is doing all the work: the rod's mass has an average $x^{2}$ of $L^{2}/12$ , the dumbbell's a flat $L^{2}/4$ .

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